How to solve this equation for c? I want to solve the equation
$$
-\frac{a}{2}\left(c+\sqrt{c^2+4}\right)=-\frac{a-1}{2}\left(c-\sqrt{c^2+4}\right)
$$
for $c$, where $a$ is just a constant.
What I get is
$$
\frac{c-\sqrt{c^2+4}}{c+\sqrt{c^2+4}}=\frac{a}{a-1}.
$$
I think there now is some "trick" to solve this for $c$.
 A: Rationalise the denominator to get $$\frac{2c^2+4-2c\sqrt{c^2+4}}{2c^2+4}=\frac a{a-1}.$$ Then after simplifying, you get $$\frac 1{1-a}=\frac c{c^2+2}\sqrt{c^2+4}.$$ If you then square both sides and make the substitution $B=1/(1-a)^2$, you should get a quadratic equation in $c^2$, which ultimately leads where you're going.
Better still, use a computer algebra system.
A: By distributing, 
$$-\frac{ac}{2}-\frac{a\sqrt{c^2+4}}{2}=-\frac{ac}{2}+\frac{a\sqrt{c^2+4}}{2}+\frac{c}{2}-\frac{\sqrt{c^2+4}}{2}$$
Well,
$$0=(a-\frac{1}{2})\sqrt{c^2+4}+\frac{c}{2}$$
$$\frac{1}{1-2a}=\frac{\sqrt{c^2+4}}{c}$$ with assumption that $c \ne0.$
Squaring, $$\frac{1}{(1-2a)^2}=\frac{c^2+4}{c^2}=1+\frac{4}{c^2}$$
$$\frac{4}{c^2}=\frac{4a-4a^2}{(1-2a)^2}$$
By fliping, and multiplying both sides by $4$,
$$c^2=\frac{(1-2a)^2}{a-a^2}$$
You now have $c$ without rationalising surds.
A: Try to isolate the square root, it is much simpler:
$$-a \sqrt{c^2+4}+a c+\sqrt{c^2+4}-c=a \sqrt{c^2+4}+a c\\
-2 a \sqrt{c^2+4}+\sqrt{c^2+4}-c=0\\
(1-2 a) \sqrt{c^2+4}-c=0\\
 \sqrt{c^2+4}=\frac{c}{1-2a}$$
Square both sides and you get:
$$c^2+4=\frac{c^2}{(1-2a)^2}$$
Rearrange that and you have a quadratic equation:
$$((1-2a)^2-1)c^2+4(1-2a)^2=0$$
Using the quadratic formula gives you:
$$c=\begin{cases}
-\frac{\sqrt{-4 a^2+4 a-1}}{\sqrt{a^2-a}}\Rightarrow -\frac{i (1-2 a)}{\sqrt{a-1} \sqrt{a}}\\
\frac{\sqrt{-4 a^2+4 a-1}}{\sqrt{a^2-a}}\Rightarrow \frac{i (1-2 a)}{\sqrt{a-1} \sqrt{a}}
\end{cases}$$
